Re: MySQL depreciated mysql_list_* commands: Replacements?

On Wed, 10 Sep 2003, Micah Stevens wrote: > describe TBL; > explain TBL; > show columns from TBL; Yes, they do give the correct information. However, my question was is there a nice simple SQL command that will return exactly what mysql_list_fields (now depreciated) so that the $result can be used with the mysql_field_* functions. All of the above commands give the exact same data in the same format, all of which are incompatible with the mysql_field_* functions. $fld_results = mysql_query("show fields from ".$db.".".$tbl) or die(mysql_error().'PMA_mysql_list_fields(' . $db . ', ' . $tbl . ')'); $fld_results_cnt = ($fld_results) ? mysql_num_fields($fld_results) : 0; for ($j = 0; $j < $fld_results_cnt; $j++) { echo mysql_field_name($fld_results, $j) . "<br/>"; } Returns: Field Type Null Key Default Extra instead of what I was expecting. > On Wed September 10 2003 12:08 pm, Peter Beckman wrote: > > So PHP is saying mysql_list_tables is depreciated. I replaced it with > > mysql_query("SHOW TABLES FROM DB") and everything is fine. > > > > Then mysql_list_fields didn't work right. So I submitted a bug > > (http://bugs.php.net/bug.php?id=25460) about it, and they said IT TOO was > > depreciated, which is fine. But I haven't figured out how to replace > > mysql_list_fields with a nice simple SQL command, since a bunch of code > > relies on mysql_list_fields to return a nice $result (ie Resource #37) to > > use with mysql_field_flags, mysql_field_name, etc. > > > > Anyone got any ideas on how to do that without replacing mysql_field_flags > > et al? Or do I have to rewrite the code entirely to use "SHOW FIELDS FROM > > DB.TABLE" and rewrite my own _field_ functions? --------------------------------------------------------------------------- Peter Beckman Internet Guy beckman@purplecow.com http://www.purplecow.com/ ---------------------------------------------------------------------------